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$xhtml = array(
	'<{title}>' => 'Learning Journal',
	'<{subtitle}>' => 'MATH 1201: College Algebra',
	'<{copyright year}>' => '2018',
	'takedown' => '2017-11-01',
	'<{body}>' => <<<END
<section id="Unit1">
	<h2>Unit 1</h2>
	<h3>Struggles with the reading assignment (the topic of the learning journal assignment)</h3>
	<p>
		The introduction for the unit simply states that a function is a set of ordered pairs in which for any <code>x</code>, there is only one <code>y</code>.
		I had to read that three times before I understood, but even after I got what it was saying, I think that statement is dead wrong.
		When y is described as a function of x, that function of x has no pairs internally, at least not necessarily.
		In most cases, the function is not a representation of ordered pairs, but a representation of some mathematical formula.
		For example, <code>f(x)</code> could be equal to <code>x + 1</code>.
		Internally, this is an equation, <strong>*not*</strong> a collection of ordered pairs.
		Ordered pairs can be <strong>*derived*</strong> from the function/equation, but that is in no way the same thing as the function/equation <strong>*being*</strong> a collection of ordered pairs.
	</p>
	<p>
		I had to read every set that used &quot;∈&quot; symbol at least twice, sometimes three times.
		The symbol&apos;s foreign to me, and I expect reading it will become easier as the course progresses if we keep using it.
	</p>
	<p>
		When the terms &quot;abscissa&quot; and &quot;ordinate&quot; were introduced, I had to read the sentence introducing them twice.
		I&apos;m unclear on why we need these words.
		Or to be more clear, I&apos;m unclear on why these words need be used.
		The terms &quot;x-coordinate&quot; and &quot;y-coordinate&quot; have the same meanings, but are much more clearly understood.
		Why were these extra terms created?
	</p>
	<p>
		I didn&apos;t have any more issues in understanding until near the end, when dealing with the difference quotient of functions.
		I had to reread the part where it said the goal is to cancel a variable from the denominator no less than six times before I understood, and even then, I had to skip ahead to the final step of the example problem solution to see what they did, then go back and continue reading.
		The problem is that this is an incredibly obtuse way of putting it.
		If the textbook had simply said we want to remove the variable from the denominator, eliminate the denominator*, or set the denominator to one, I&apos;d&apos;ve understood right away without any issues.
		(*I&apos;m aware you can&apos;t technically eliminate the denominator; there&apos;s always a denominator.
		Technically, you&apos;re only setting the denominator to one.
		Still, saying to eliminate the denominator gets the message across.)
	</p>
	<p>
		I had to read step three of the music player example so many times I lost count.
		The issue was that I was looking at the problem from the wrong angle.
		I forgot that the price function was given by the problem and not at all based on factors such as production costs.
		It didn&apos;t help that the given price function makes no sense.
		I mean, it gives a price charged each for zero units.
		If you produce and sell nothing, the price per unit is undefined, so with a price function that magically defines it anyway, there&apos;s bound to be confusion.
		I thought I figured out what was going on, but really hadn&apos;t, so I likewise struggled on step four.
		In trying to make concrete my explanation on why I was so confused on these two steps, I managed to fix my understanding of step three, and in doing so, completely removed my confusion on step four.
	</p>
	<h3>Other notes</h3>
	<p>
		I didn&apos;t struggle with grasping this concept, but it&apos;s worth mentioning that it&apos;s dead wrong: the book says we can&apos;t use the Pythagorean Theorem to find the distance between two points directly aligned either vertically or horizontally.
		We can quickly see that it would be overkill to use the Pythagorean Theorem, and it&apos;d certainly be a waste of resources if performing the calculation by hand, but it&apos;s very much an option.
		The key to doing so lies in the definition of what a triangle is.
		Have you heard of <a href="https://en.wikipedia.org/wiki/Degeneracy_%28mathematics%29#Triangle">degenerate triangles</a>?
		I actually had to bring those up last term as well because of a <a href="https://y.st./en/coursework/CS2401/Triangles.xhtml">strange assignment</a>.
		In any case, the Pythagorean Theorem functions perfectly well and exactly as expected, even if the points share one of their two coordinates (or even both coordinates).
		As mathematicians, we should understand such concepts as degeneracy.
		Degeneracy allows us to realise that points that share a coordinate or two are not a special case when it comes to distance.
		Additionally, I&apos;m working on my computer science degree right now.
		Understanding degeneracy allows me to avoid needlessly complicating my code if I were to write software for calculating such distances because I don&apos;t have to account for special cases where no special cases actually exist.
		Both as a mathematician and as a programmer, this seems to me like a basic and fundamental concept that shouldn&apos;t be overlooked in favour of a special-cases-based view of the situation.
		Saying an easier method for calculating such cases exists is fine, but claiming that the standard approach somehow doesn&apos;t work is deceitful and misleading.
	</p>
</section>
<section id="Unit2">
	<h2>Unit 2</h2>
	<p>
		Perhaps I misunderstood the material last week.
		This week, the textbook tells us the graph of function f() is a graph of all the points in which y = f(x).
		But ... isn&apos;t that what it told us last week?
		I mean, it didn&apos;t tell us in those works, but it claimed that the y values were the output from such a function.
		The phrasing this week makes it out like this is a new concept to us though; that we haven&apos;t already covered it in past sections.
	</p>
	<p>
		It occurred to me while reading this week that there can be any non-negative integer of x-intercepts, but there can only be zero or one y-intercept.
		I guess this should be rather obvious, but it had never occurred to me before this week, even in past maths courses at other schools.
	</p>
	<p>
		I struggled in reading the example for the piecewise function.
		I understood completely how to graph such a function before I started reading this week, that wasn&apos;t even the issue.
		I just didn&apos;t get the phrasing, so I had to reread parts of that paragraph repeatedly.
	</p>
	<p>
		The book asks why we&apos;re dismissing symmetry across the x-axis, leaving it to the reader to figure it out.
		Even before reaching that question, I&apos;d already had the answer.
		Despite my struggles, even I can figure out that symmetry about the x-axis results in something that doesn&apos;t fit the definition of a function; y would have to have two values for each value of x.
		Aside from values in which y = 0, of course.
		Which brings me to the next question presented by the book: what functions could be both even and odd?
		This probably doesn&apos;t cover all even-and-odd functions, but the function f(x) = 0 is both even and odd.
		We can derive other functions from this by limiting the domain.
		The domain can be -a to a, for all values of a, and the function will retain both symmetries, as well as the third symmetry (about the x-axis), also shared by the bare f(x) = y function.
	</p>
	<p>
		When we look at the formal definitions for a local maximum and local minimum, we see (unintuitively), that all points on a constant interval besides the endpoints fit the definition of both.
		That is to say, on a constant interval, you have an infinite number of local maximums that are also local minimums.
		Isn&apos;t that bizarre?
	</p>
	<p>
		I had to reread <q cite="http://www.stitz-zeager.com/sprecalc07152011.pdf">Since g(x) = (x + 2) 2 − 3 = f(x + 2) − 3</q> countless times by the end of the week.
		I just couldn&apos;t understand where it was getting that from.
		I had to have read it at least a dozen and a half times in total, though I lost track of the exact number.
		Strangely, the book doesn&apos;t cover the easy plug-and-solve method for finding vertices; there&apos;s actually an equation for that.
		Or rather, there&apos;s an equation for finding the axis of symmetry.
		Once you have the axis, you have your x value, and with the x value, you can plug it into the parabola&apos;s function to get the y value.
		Instead, the book only covers how to find particular points on the parabola using transformations from the most basic parabola.
		If you apply a transformation to the vertex point, you can find the vertex.
		That method is much more obtuse, but once you finally wrap your head around it, it works.
		(Maybe it&apos;s just me that&apos;s struggling with this though.
		Then again, the posts in the discussion forum tell me there are students in worse shape than I am, failing to grasp the concept that January and February are two spans on the same continuum of time, and that lines drawn for each span can&apos;t have the same x values; that they can&apos;t &quot;intersect&quot;.
		I&apos;m not the only one not comprehending right away what should be basic concepts.)
	</p>
	<p>
		I ended up having to do the unit assignment before I finished the reading assignment, due to a shortage of time.
		Honestly, that was my own fault.
		Anyway, I had to figure out transformations before reading all of what the book had to say about them.
		In particular, vertical horizontal transformations confused me.
		I could understand how to use them, but I didn&apos;t understand the maths behind why they work the way they do.
		The textbook gave a pretty good explanation though and cleared things up.
		It&apos;s actually pretty reasonable, they way they explained it.
	</p>
	<p>
		After reading about both horizontal transformations and horizontal scaling, I realised that doing anything horizontally is likely to have the opposite of the intuitive effect.
		When you multiply by two, the graph gets <strong>*thinner*</strong>.
		When you add four, the graph moves to the <strong>*left*</strong>.
		This is because numbers work in reverse in the input.
		Changing the inputs in the new function means that within the new function, the reverse operation must be applied before &quot;calling&quot; the old function.
		This isn&apos;t a programming course, but I&apos;m afraid I&apos;m a programmer.
		The best way I can explain this is in a code snipet.
		If g(x) = f(x+2), then g(x) is basically doing this:
	</p>
	<blockquote>
<pre><code>function g(\$x) {
	return f(\$x-2);
}</code></pre>
	</blockquote>
	<p>
		As we can see, the input to the original f(x) is transformed in the reverse way we expect it to, resulting in a reversed effect.
		The same applied to multiplication, division, exponents, et cetera.
		The g(x) will always be reversing our transformation equation before inputting it into f(x).
		A few paragraphs after figuring that out, the book pretty much told me the same thing.
	</p>
	<p>
		I&apos;m confused as to the purpose of the point-slope form of the equation of a line.
		It looks like an unreduced equation, to me.
		Meanwhile, the slope-intercept form described next perfectly describes y in terms of x.
		There&apos;s a reason that&apos;s the common form to use.
	</p>
	<p>
		It took a while to understand the proof on the fact that all quadratic equations can be written in standard form.
		And then, right there after the proof, was the vertex formula I&apos;d needed for the assignment.
		I&apos;d looked that up elsewhere, but then didn&apos;t use it because I hadn&apos;t reached it in the book and thought the only method in the book was the one involving transformations.
		In the coming week, I need to make sure to find the time to read the entire reading assignment as early on as possible.
	</p>
</section>
<section id="Unit3">
	<h2>Unit 3</h2>
	<p>
		I enjoyed the real-world example, though it did take a couple times going back to verify why all the numbers were used as they were.
	</p>
	<p>
		The sections on the end behaviour of functions took heavy rereading as well.
		If put in plain English, I doubt I&apos;d have to reread them even once, but they&apos;re instead packed full of symbols.
		I guess that&apos;s for &quot;clarity&quot; or something, but it makes it harder to take in the message they&apos;re trying to convey.
	</p>
	<p>
		The textbook says many students find the intermediate value theorem to be silly.
		I would be one of those students.
		Like the book says, this is just basic logic; it&apos;s not really some big breakthrough principle.
	</p>
	<p>
		I understood the theorem of the end behaviour of polynomials right away, and it seemed intuitively like it should be valid.
		After all, the higher the exponent, the more extreme the behaviour graphically.
		High exponents give us quickly-increasing, steep slopes.
		Their effect on the line should, in theory, dwarf all terms with smaller exponents in the same equation, especially at high values of x.
		However, the book quickly lost me in its proof of the theorem.
		It rewrote the equation of a given line into a bizarre form, with no explanation as to why or as to how it got to the new equation.
		At first, the new equation didn&apos;t even seem equivalent to the original, though after a bit, I was able to figure out what the book had done and what it was trying to accomplish.
		After finally getting what was going on, I continued, at which point the book explained in the <strong>*section on the theorem*</strong> that it was factoring out the leading term.
		Couldn&apos;t it have <strong>*started*</strong> with that information instead of just calling the initial factoring a &quot;rewriting of the equation&quot;?
		It would have made things much easier to grasp and would&apos;ve saved me a lot of frustration.
	</p>
	<p>
		I didn&apos;t understand hardly any of the beginning of section 3.2.
		However, it was then brought up that what we&apos;d just gone over was to establish the connection between the factors of a polynomial and the zeros of that same polynomial.
		The connection is already clear to me: once a polynomial is factored out, setting x to any value that zeros out one of the factors results in that value of x being a zero of the function.
		The reason for this is that it causes the rest of the equation to be multiplied by zero, making said rest of the equation irrelevant.
		There are probably some subtleties I&apos;m missing because I don&apos;t understand the provided explanations, theorems, and proofs, but hopefully understanding the concept and the basic logic behind the concept will be enough.
	</p>
	<p>
		I am so confused as to where this &quot;synthetic division&quot; comes from.
		I can work with it to solve problems, but I don&apos;t at all understand how it even functions properly.
		The explanation given by the book wasn&apos;t at all clear on this matter.
		It certainly took the time to explain, but the explanation didn&apos;t seem to make much sense.
		Either I&apos;m inept, or the explanation was geared toward people of a higher mathematical skill level, not for beginners in this type of maths.
	</p>
	<p>
		The book tries to explain why you can&apos;t use i to make 2 equal to -2, but I don&apos;t get where it&apos;s coming from.
		I mean, I completely agree that 2 ≠ -2, but what exactly is it trying to disprove?
		In the example, negative four is replaced with negative negative four (positive four).
		Of <strong>*course*</strong> you&apos;re going to end up with a different solution when you throw an extra negative in there.
		I can throw an extra negative into just about any equation and make it spit out a different number; that obviously doesn&apos;t prove that the original solution and the solution to the modified problem are somehow equal.
	</p>
	<p>
		Either the book didn&apos;t explain how to decide what to divide by when looking for zeros or I somehow completely missed it.
		In either case, the example presented right after the complex number theorem baffled me.
		Reaching the solution involved dividing by (x - ½), but no explanation of where the ½ came from was given.
		Trying to figure out the solution myself, I eventually decided to try to work backwards.
		I knew the far right term had to cancel out the one above it, so I&apos;d try manipulating the numbers to make that happen.
		As I did so, I stumbled upon the missing information.
		I needed a negative one, and the term before was a negative two, if unmodified.
		Multiplying by a half resulted in the necessary term, and that could be cascaded upward, instead of downward.
		I tried that on the homework though, and it didn&apos;t work out as expected.
		Whatever was done, it wasn&apos;t that.
		I went back to reread, and I found the issue: the instructions are in section 3.3, but our reading assignment is sections 3.1, 3.2, and 3.4.
		We skip section 3.3 this week, but section 3.3 is kind of a big deal for solving the homework for the week.
		Whose idea was it to skip that section!?
	</p>
</section>
<section id="Unit4">
	<h2>Unit 4</h2>
	<p>
		The section explaining the &quot;inside out&quot; and &quot;outside in&quot; methods of composing functions with one another was incredibly confusing.
		I had to reread the first example showing the difference several times before I could figure out what they were doing.
		It turns out it&apos;s because the book was starting a step back from where I start, which is probably helpful for people having a harder time grasping the concept of function composition.
		In both proposed methods, only step zero differs.
		From step one on, everything is the same.
		But their step one is my step zero.
		I just sort of ... merge the functions, if that makes any sense.
		I take the one function&apos;s internal equation and insert it into the internal equation of the other function.
		The book, on the other hand, starts by putting the function &quot;call&quot; (to use a computer science metaphor) of the inner function into the equation of the outer function or puts the function &quot;call&quot; of the outer function around the equation of the inner function.
		It&apos;s extra effort with (if you know what you&apos;re doing) no added benefit.
	</p>
	<p>
		The book states that composing a function with itself seems odd, but is perfectly acceptable.
		It doesn&apos;t really seem that odd though.
		It&apos;s like any other mathematical operation.
		For example, is it odd to add 2 + 2, or subtract five from itself?
		Not really.
		Then the book called such compositions &quot;iteration&quot;.
		Nice.
		More parallels with computer science terminology.
		That&apos;s not <strong>*quite*</strong> how iteration works in programming, but still ... it&apos;s kind of similar.
	</p>
	<p>
		I spent <strong>*way*</strong> too much time on one of the quiz questions:
	</p>
	<blockquote>
		<p>
			For f(x) = √(x + 5) + 2, what is the domain of f<sup>-1</sup>?
		</p>
	</blockquote>
	<p>
		I tried to solve this without writing (er, typing) anything down.
		I kept trying to keep the inverted function in my head so I could examine the range, and I kept doing things wrong and having to start again.
		But then I realised something very basic, that&apos;s a bit embarrassing not to have noticed from the start: the domain and range of the inverse are equal to the range and domain of the original function, respectively.
		I don&apos;t need to know the equation of the inverted function.
		I can forget the inverted function entirely.
		All I needed to do was report the range of the original function.
		That was a <strong>*much*</strong> more manageable task, and only took a second or two.
		I guess I was just too tired to think of that right away when I took the test.
	</p>
	<p>
		This week&apos;s reading material was pretty easy, so I didn&apos;t really have a whole lot of struggles with it.
		Part of that is because the assigned reading material was so short, but also, I work with functions in programming so often.
		Computer science functions are very much like mathematical functions, especially when the function inputs are numeric.
		Some people don&apos;t use algebra much, but as a computer scientist, I actually do maths like this for <strong>*fun*</strong>.
		Just recently, I built a graphical front end for a stat counter in a game I play, using algebra to place the items at the correct screen coordinates.
		Longer ago, I built the stat counter itself, taking all sorts of numeric variables into account (among other things).
	</p>
</section>
<section id="Unit5">
	<h2>Unit 5</h2>
	<p>
		The book used the word <a href="https://en.wikipedia.org/wiki/Asymptote">asymptote</a>, but never explained it.
		Thankfully, the Internet usually has answers for questions of simple definitions.
	</p>
	<p>
		The section explaining why we can&apos;t use a negative number as a base was a bit confusing.
		Part of it I think is that it seems to expect us to just know what fractional exponents do.
		We didn&apos;t study that though.
		I&apos;m not quite clear on how they work.
		From what I gather from the surrounding text though, their denominators are used to determine a root we need to calculate on the base.
		For example, for x<sup>1/2</sup>, we&apos;d find the square root of x, for x<sup>1/3</sup>, we&apos;d find the cube root of x, et cetera.
		That doesn&apos;t explain the whole story though; a fraction has more than a denominator.
		x<sup>1/3</sup> should obviously not be equal to x<sup>2/3</sup>!
		Maybe we multiply the root by the numerator?
		I don&apos;t know; it&apos;s beyond the scope of what we&apos;re studying at the moment.
		Anyway ... the fact that when x is fractional with an even denominator, we can&apos;t find real solutions to (-2)<sup>x</sup>, makes me wonder what the real segments of such a line would be.
		It seems like the line would drift between the real and imaginary number planes as the denominator of x changes.
		Furthermore, there are an infinite number of irrational numbers between each discernible, rational fraction.
		At what points does the line switch planes?
		An irrational number cannot be expressed as a proper fraction, so it can&apos;t have an even or odd denominator.
		How do we know which values of x allow the equation to be solved?
		I guess that&apos;s, again, outside the scope of what we&apos;re studying right now though.
		Maybe we&apos;ll cover this sort of thing in <span title="Calculus">MATH 1211</span>, <span title="Introduction to Statistics">MATH 1280</span>, or <span title="Discrete Mathematics">MATH 1302</span>.
	</p>
	<p>
		The rest of the text this week was easy to understand.
		Logarithms are a bit strange, but their application makes sense.
	</p>
	<p>
		I did find I hadn&apos;t quite understood the part on radioactive decay once I got to the written assignment though, so I had to go back and try to figure that out.
		When I used the equation the way I thought I was supposed to, I got an answer that made zero sense whatsoever.
		I forget my exact result, but the missing carbon seemed to predict that the object containing it was less than a year old.
		I managed to find the issue though, which is that we&apos;re supposed to use the equation <strong>*twice*</strong>.
		The first time, we use the half life to determine a constant k, then the second time, we use that constant k to figure out the age of the object.
		But why?
		In working my way through the problem the book&apos;s way, I managed to piece together a new equation that combines the two steps:
	</p>
	<p>
		A(t) = A<sub>0</sub>e<sup>ln(½)÷h*t</sup>
	</p>
	<p>
		Here, t is the time as before, while h is the half-life.
		It works by substituting the results of the first round of the equation for k, so we don&apos;t even need to bother with k any more.
		I mean, what even <strong>*is*</strong> k?
		As far as we were told, this constant only has application when computing radioactive decay, but it&apos;s not this constant that people keep track of.
		It&apos;s the half life.
		So why not just use the half life directly like this?
		I actually only stumbled upon that new equation because I was carrying over exact results instead of using a calculator to estimate before plugging into the new equation, but surely I can&apos;t be the only person to have figured this out.
	</p>
</section>
<section id="Unit6">
	<h2>Unit 6</h2>
	<p>
		The initial introduction to parametric solution notation threw me off for a bit.
		Until I got to the end of the explanation, I had no idea why the author was doing what they were doing.
		It did make sense by the time the explanation was complete though; no explanation as to why was ever given, but it seems like a reasonable notation to use.
		It&apos;s concise enough too, though it takes a while to work a problem into that format of solution.
		Once in that format, it&apos;s also very easy to work out individual solutions by simply plugging in any possible value.
	</p>
	<p>
		I also found it surprising that we are covering systems with no solutions, and that there&apos;s even a name for them: inconsistent systems.
		I certainly already knew that one could design sets of equations in which the values of the variables don&apos;t allow for a solution; just throw random numbers together with your variables in multiple equations and you&apos;ll more than likely end up with something unsolvable.
		However it hadn&apos;t occurred to me that calling such arrangements systems and actually taking them seriously could have applications.
		I guess it makes plenty of sense though.
		As a programmer, I use plenty of algebra in my code.
		When I&apos;m working with game-related code, I sometimes end up trying to engineer an equation that will give me a particular solution at a particular point, or I try to find an intersection of two or more equations.
		Sometimes, there is no intersection and I have to figure out new equations.
		These inconsistent systems we&apos;re studying now are an abstract form of something I&apos;ve already dealt with before.
		Such systems clearly have application in the real world, though it doesn&apos;t seem at first glance that they would.
		I&apos;m lucky enough to be in a position to see such a case, though I&apos;m not sure many other students taking the course are so fortunate.
		It&apos;s not a case that shows up in a great many fields, I&apos;d wager.
		Then again, programming has more to do with maths than many fields.
	</p>
	<p>
		The numbering of the cells of matrices is incredibly unintuitive.
		Throughout the textbook, we&apos;ve always numbered from zero.
		Variables in definitions, for example, would often be a<sub>0</sub>, a<sub>1</sub>, a<sub>2</sub>, et cetera.
		Matrices are introduced as being instead indexed from one.
		Not only is this needlessly inconsistent, but as a programmer, I much prefer indexing from zero regardless of context.
		Zero is actually a pretty important number, and was a huge development in mathematics.
		We didn&apos;t always have a concept of zero, and it&apos;s a shame to just skip it as if we still don&apos;t have it to this day when numbering things.
	</p>
	<p>
		For the most part though, this week&apos;s material was pretty straightforward and easy to grasp.
		I was worried at the beginning of the term because I wasn&apos;t getting certain things; I thought the course might be above my level.
		it seems it was just bad luck or something in the beginning.
		Everything else is coming together just fine.
	</p>
</section>
<section id="Unit7">
	<h2>Unit 7</h2>
	<p>
		I found it interesting that the first section of the reading material this week, which was on angles, jumped right to the case of degenerate angles, though it didn&apos;t refer to them as such.
		They are the case of the 180° angle and the 0° angle.
		To the layman, these certainly don&apos;t look like angles, but like degenerate triangles are to the general case of triangles, they very much are from a mathematical perspective, as they fit into the definition of an angle and follow all the same patterns as other angles.
	</p>
	<p>
		The example for finding the complementary angle for an angle with a size formatted as degrees, minutes, and seconds threw me off for a bit.
		The author chose to reformat 90° as 89°59&apos;60&quot;, which was bizarre.
		I guess it works if you don&apos;t know how to properly work across the segments of that value and &quot;borrow&quot; as you need to, but this is an extra, useless step for the rest of us.
		I have no doubt the author knows how to do it without the rewrite, but provided the rewrite for the ease of students less versed in maths.
	</p>
	<p>
		The book next introduces oriented angles, which are basically signed angles, or angles with a positive or negative sign.
		(I&apos;m a programmer, so I deal with signed numbers, which can be positive negative, or zero; and unsigned numbers, which can only be non-negative.
		Hence, my use of the term &quot;signed&quot; in regards to angle measures.)
		For years, I&apos;ve wondered why people never measure angles that way, though I guess what I came up with was very slightly different.
		It&apos;s nice to know people actually use this concept though.
		It&apos;s strange that measuring in the counter-clockwise direction is positive, not negative, but I can certainly work with that.
		Greater-than-360° angles are also easy to understand, and quite remind me of my understanding of modulo division.
		Many people consider modulo division to be the finding of some &quot;remainder&quot;, but given that we&apos;re not limited to integer values in our answer, there <strong>*isn&apos;t*</strong> a remainder when we divide.
		Likewise, we can use a fractional value when performing modulo division (3.5 modulo 2 is 1.5, for example), but the concept of a &quot;remainder&quot; entirely falls apart when fractional values are used.
		Instead, I see modulo division as cycling around like this.
		In computer science, we call this overflow.
		You can also think of it like a clock that starts over when it reaches the end, or like an odometer, cycling back around past its maximum value.
		In the case of these large angles, because they cycled past the &quot;maximum&quot; value, they look smaller than they actually are, and I&apos;m guessing that if we were to use this sort of thing in an equation, it&apos;d be for physics and we&apos;d perform that modulo division to get a &quot;proper&quot; angle measurement, for lack of a better term.
		The book even basically uses modulo division in comparing coterminal angles.
	</p>
	<p>
		It took me a bit to work out the language used to explain what radians are though.
		You can&apos;t blame me though; the definition is quite strange.
		Degrees are easier to work with because of the typically-nicer numbers.
		I mean, a full circle measured in degrees yields a rational number.
		However, the definition of a degree is, as far as I know, completely arbitrary.
		When you derive your measurement unit from comparing the radius to the perimeter ... at least your unit is coming from <strong>*somewhere*</strong>.
		I can see why some people would prefer radians.
		However, the book continues that this is useful because we can wrap a number line along a unit circle and the measure of the angle in radians will match the length of the ark formed in units.
		No explanation as to how this is a trait that even <strong>*resembles*</strong> something useful is given.
		I&apos;m thoroughly confused.
	</p>
	<p>
		I didn&apos;t understand the section on circular motion at first, as I mistakenly thought it was referring to the motion of spinning.
		That&apos;s not what it meant at all.
		It&apos;s talking about an arc-shaped trajectory of movement.
	</p>
	<p>
		When it was discussed that we could use reflection as a shortcut for finding the sine and cosine of a point on the unit circle, I had to go back ad reread that a couple times.
		I guess it&apos;s usable, but it assumes we have the information on the reflected point.
		If we&apos;re solving a problem that involves multiple points, we may very well have this information already, making this shortcut viable.
		However, needing this information is a very limiting factor and we should keep that in mind.
	</p>
	<p>
		<del>I&apos;m confused by extended interval notation.
		Supposedly, it records all the holes in the domain, but I can&apos;t see how or where.
		It looks, from that notation, to be a continuous, unbroken line.</del>
		<ins>Actually, scratch that, I figured it out.
		I was reading the parenthesis as delimiters for an ordered pair, but they&apos;re being used as delimiters for an interval.
		The use of the same symbols for multiple purposes makes it harder to pick up the new concepts.</ins>
	</p>
</section>
<section id="Unit8">
	<h2>Unit 8</h2>
	<h3>Problem 0</h3>
	<p>
		The sin() function has a range of [-1, 1].
		If we&apos;re adding two to the result of it to get one, logically, we&apos;re actually looking for where sin() is returning the value at the opposite end of its range, negative one.
		To make better sense of the equation, let&apos;s subtract two from both sides:
	</p>
	<p>
		sin(x - π) + 2 = 1<br/>
		sin(x - π) = -1
	</p>
	<p>
		Next, if we imagine the sin() function as reporting values on the unit circle at a given arc length, subtracting pi flips our answers halfway around the circle.
		I&apos;m honestly not sure how to express that concept directly in a mathematical way, but thankfully, I don&apos;t have to.
		There&apos;s another way.
		We can use the sin() subtraction identity.
	</p>
	<p>
		sin(x - π) = sin(x)cos(π) - cos(x)sin(π) and sin(x - π) = -1
	</p>
	<p>
		So:
	</p>
	<p>
		sin(x)cos(π) - cos(x)sin(π) = -1
	</p>
	<p>
		Going back to our unit circle visualisation, the sin() function reports y values while the cos() function reports x values.
		An arc length of π puts our terminal edge halfway around the unit circle, at the point (-1, 0), so:
	</p>
	<p>
		sin(π) = 0 and cos(π) = -1
	</p>
	<p>
		Using that information, we can reduce our equation.
		We don&apos;t even need to deal with cos(x), as it gets multiplied by zero.
		We thus get:
	</p>
	<p>
		sin(x)cos(π) - cos(x)sin(π) = -1<br/>
		-1sin(x) - 0cos(x) = -1<br/>
		-sin(x) = -1<br/>
		sin(x) = 1
	</p>
	<p>
		We&apos;re pretty much right back where we began.
		Subtracting pi from sin()&apos;s input flipped us halfway around the circle, while adding two to the output flipped us the rest of the way, putting us at the beginning.
		We&apos;ve just found that:
	</p>
	<p>
		sin(x - π) + 2 = sin(x)
	</p>
	<p>
		With that in mind, all we need to do now is look at our unit circle once more to determine where in the range specified by the problem, [0, 2π), sin(x) is equal to one.
		If we remember that sin() is reporting y values on the unit circle, the answer becomes obvious.
		At only one point does the unit circle&apos;s y value ever get as high as one, and that&apos;s the point one quarter of the way from the beginning. Because the perimeter is 2π, we divide that value by four and find that the only answer within the specified range is <strong>0.5π</strong>, which can alternatively be written as <strong>π÷2</strong> (pi over two).
	</p>
	<h3>Problem 1</h3>
	<p>
		The arcsin() function is designed to be the inverse of the sin() function.
		However, the sin() function, when graphed, produces a wave-like form, oscillating up and down.
		As it&apos;s continuous, what this tells us is that the same y value can be produced from multiple x values.
		In other words, because sin() is a cycle, it doesn&apos;t pass the horizontal line test, meaning that the sin() function is <strong>*not*</strong> invertible.
		Arcsin() is the &quot;inverse&quot; of an irreversible function.
		To accomplish its task, it looks at only a small section of the sin() function&apos;s domain and uses that as its own range.
		If our original number, before using sin() and arcsin(), is outside this small section of sin()&apos;s domain, we&apos;ll get back a different value from arcsin() than we passed to sin().
		It can&apos;t be helped.
	</p>
	<p>
		A good way to think of this is that it&apos;s similar to lossy compression.
		If you take a $a[PNG] file and convert it into a $a[JPEG], data is lost.
		The same happens when you find the sine of a number.
		Taking the arcus sine of the returned value is like trying to convert the $a[JPEG] back into a $a[PNG].
		The data&apos;s already been lost.
		You can&apos;t get it back.
		Instead, you get something that shares qualities with the original, but it&apos;s not (necessarily) quite the original itself.
	</p>
	<h3>Problem 2</h3>
	<p>
		This problem asks what the four values x can be in the following equation:
	</p>
	<p>
		x<sup>4</sup> = 1
	</p>
	<p>
		I have no idea at all how to show work on this; all I can do is report what intuitively works and explain a little why they make the equation true.
		<strong>Our possible x values are 1, -1, i, and -i.</strong>
	</p>
	<h4>One</h4>
	<p>
		One to any power should go without explanation as being equal to one.
		One is the multiplication identity, so it doesn&apos;t matter how many times you multiply it by itself, it&apos;ll always remain just one.
	</p>
	<h4>Negative one</h4>
	<p>
		For negative one, we rely on the fact that multiplying a negative by a negative results in a positive number.
		We can split the problem up a bit.
		Instead of looking at -1*-1*-1*-1, we&apos;ll look at (-1*-1)(-1*-1):
	</p>
	<p>
		-1<sup>4</sup> = 1<br/>
		(-1<sup>2</sup>)<sup>2</sup> = 1<br/>
		1<sup>2</sup> = 1
	</p>
	<p>
		After breaking it up and multiplying away our negative, we&apos;re left with 1<sup>2</sup>.
		Like with the last solution, we&apos;re left with an identity multiplied by itself, which will always equal itself.
	</p>
	<h4>The square root of negative one</h4>
	<p>
		This solution is a bit strange, but we can break it down like in our last solution, multiplying (i*i)(i*i).
		By definition, i multiplied by itself equals negative one.
		From there, we have negative one multiplied by negative one, which yields one:
	</p>
	<p>
		i<sup>4</sup> = 1<br/>
		(i<sup>2</sup>)<sup>2</sup> = 1<br/>
		-1<sup>2</sup> = 1<br/>
		1 = 1
	</p>
	<h4>The negative square root of negative one</h4>
	<p>
		This is our most complicated solution.
		(Or should I say that it&apos;s our &quot;most complex&quot; solution?)
		For clarity, let&apos;s not work with -i, but instead -1i.
		That lets us better see that the negative sign itself can be multiplied by itself in the form of -1.
		So basically, we begin working like this:
	</p>
	<p>
		-i<sup>4</sup> = 1<br/>
		(-1i)<sup>4</sup> = 1<br/>
		((-1i)<sup>2</sup>)<sup>2</sup> = 1<br/>
		((-1i)(-1i))<sup>2</sup> = 1<br/>
		(-1 * i * -1 * i)<sup>2</sup> = 1
	</p>
	<p>
		Here, we see our inner expression involves two instances of i and two instances of negative one being multiplied together.
		When dealing with several numbers being multiplied together, it doesn&apos;t matter which order we perform the multiplication; it&apos;ll come out the same every time.
		Additionally, we can even multiply terms that aren&apos;t next to each other in the problem, so long as they&apos;re part of the same multiplication chain.
		For that reason, we can rearrange the terms:
	</p>
	<p>
		(-1 * i * -1 * i)<sup>2</sup> = 1<br/>
		(-1 * -1 * i * i))<sup>2</sup> = 1<br/>
		(-1<sup>2</sup> * i<sup>2</sup>))<sup>2</sup> = 1
	</p>
	<p>
		Squaring negative one results in one, while by definition, squaring i results in negative one.
		With one being the multiplication identity, it basically just dissolves on us, cancelling itself out, leaving us with negative one squared, which we just said is equal to one:
	</p>
	<p>
		(-1<sup>2</sup> * i<sup>2</sup>))<sup>2</sup> = 1<br/>
		(1 * -1)<sup>2</sup> = 1<br/>
		-1<sup>2</sup> = 1<br/>
		1 = 1
	</p>
	<p>
		There&apos;s probably a <strong>*way*</strong> easier way to justify -i being a solution to the problem, but I don&apos;t currently know how to go about that.
	</p>
	<h3>Difficulties on the reading assignment</h3>
	<p>
		The book tried to explain function arguments as being the thing &quot;inside&quot; a function, which is a highly unintuitive way to think about it.
		The thing inside the function is the stuff that makes it tick.
		It&apos;s the guts of the function, the very thing that makes the function do as it&apos;s defined to do.
		The argument, on the other hand, is what you <strong>*feed into*</strong> the function.
		It&apos;s not inside the function, as it&apos;s not even a part of the function.
		I guess you could say the argument is inside the function <strong>*call*</strong>, but it&apos;s not inside the <strong>*function*</strong>.
		This is a very important distinction when trying to understand what a function is and how to use one.
	</p>
	<p>
		The book claimed that the sec() and csc() functions are the reciprocals of the cos() and sin() functions, which confused me for a bit.
		This statement by the book is actually <strong>*not true*</strong> though.
		They&apos;re an approximation of reciprocals, but to be true reciprocals, multiplying them by the respective functions they&apos;re reciprocals of would need to result in a continuous function.
		Because sin() and cos() include zero in their ranges, they can&apos;t have true reciprocals, which is why the csc() and sec() functions have holes in their domains.
	</p>
	<p>
		It confused me at first that the domain segment taken from the sin() and cos() functions to create the arcsin() and arccos() functions didn&apos;t match.
		The reason for this was never explained and seemed at first to be arbitrary.
		It&apos;s actually about getting a segment that doesn&apos;t double back on itself in terms of y.
		If it doubles back, it won&apos;t pass the horizontal line test and can&apos;t be inverted.
	</p>
	<p>
		The concept of the θr plane threw me off for a bit.
		I figured out what it was talking about after a bit, but it isn&apos;t a plane we&apos;ve worked with before.
		It didn&apos;t seem to make sense that r wasn&apos;t remaining stable either, but I hadn&apos;t accounted for the fact that r didn&apos;t refer to the distance from the centre of the circle; it referred to the distance from the origin, which the circle wasn&apos;t centred on.
	</p>
	<p>
		Much of the material relating to moduli in the polar coordinate system was made very difficult to read by the reuse of symbols.
		|z| was used to indicate the modulus, but I was finding it incredibly hard not to read that as &quot;the absolute value of zed&quot;.
		As the modulus is always the positive distance from the origin, that made things even more difficult to keep straight, as it seems very much like it <strong>*is*</strong> the absolute value.
		There&apos;s a huge difference though.
		z in this case is being used as a coordinate pair, not the single coordinate.
		Expressed in the polar coordinate system, we have this:
		z = (r, θ).
		In other words, |z| is <strong>*not*</strong> the absolute value of z, and the concept of an absolute value of z doesn&apos;t even make any sense, as far as I know.
		Rather, |z| is the absolute value of <strong>*r*</strong>!
		Very unintuitive.
		From there, we started multiplying things together within the pipe characters, using terms such as |zw|.
		That didn&apos;t seem right at first, as you can&apos;t multiply coordinate pairs together.
		I had to read that a few times to remember that z and w are also complex numbers, not just coordinate pairs.
	</p>
	<p>
		I didn&apos;t really understand the product, power, and quotient rules for complex numbers expressed in polar form at all.
		By the time I&apos;d gotten to that part, I was already so lost, and those rules built on what we&apos;d previously covered.
		From this point on, I felt like I was pretty much staring at a mess of symbols.
		None of it was really making sense.
		I tried finding other resources on the Internet to help, but I wasn&apos;t getting anything useful from the sources I found, either.
	</p>
</section>
END
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